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Numerical Ability Test - 3 - Bank Exams MCQ


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30 Questions MCQ Test - Numerical Ability Test - 3

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Numerical Ability Test - 3 - Question 1

What will come in the place of the question mark '?' in the following question?
(17 × 84) + (20% of 960) = (16 × 24) + (? × 12)

Detailed Solution for Numerical Ability Test - 3 - Question 1

Follow the BODMAS rule to solve this question, as per the question is given below:

Step 1: Parts of an equation enclosed in ' Brackets ' must be solved first and in the bracket,

Step 2: Any mathematical ' of ' or 'exponent' must be solved next,

Step 3: Next, The part of the equation that contains 'Division' and 'multiplication' are calculated,

Step 4: Last but not least, the parts of the equation that contains 'Addition ' and 'Subtraction' should be calculated

Given :

⇒ (17 × 84 ) + (20% of 960 ) = (16 × 24 ) + (? × 12)

⇒ 1428 + 20 × 960 / 100 = 384 + ? × 12

⇒ 1428 + 192 - 384 = ? × 12

⇒ 1620 - 384 = ? × 12

⇒ 1236 = ? × 12

⇒ ? = 1236 /12

⇒ ? = 103

Hence 103 is the correct answer.

Numerical Ability Test - 3 - Question 2

What should come in the place of the question mark '?' in the following question?

Detailed Solution for Numerical Ability Test - 3 - Question 2

Given:

Follow BODMAS rule to solve this question, as per the order given below:

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Now,

Calculation:

√? = 25

? = 252

? = 625

The correct answer is 625.

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Numerical Ability Test - 3 - Question 3

Aditya sells goods at 10% loss on cost price but uses 20% less weight. Find the profit or loss percent.

Detailed Solution for Numerical Ability Test - 3 - Question 3

Given:

Aditya sells goods at 10% loss

He uses 20% less weight

Formula used:

Gain% = (Gain/C.P) × 100

S.P = C.P × (100 – loss%)/100

Calculation:

Let the cost price of 1000 gm be Rs. 100

Loss% = 10%

S.P = C.P × (100 – loss%)/100

So, S.P = 100 × (90/100)

⇒ S.P = Rs. 90

Now, he is using weight 20% less

So, weight he is using = 1000 × (1000 – 20)/100

⇒ Weight he is using = 800 gm

So, the cost price will be on 800 gm only

So, Total cost price of 800 gm = Rs. 100 × (800/1000)

⇒ Rs. 80

And, total selling price of 1000 gm = Rs. 90 × (1000/1000)

⇒ Rs. 90

Gain = S.P – C.P

⇒ Gain = Rs. 90 – Rs. 80

⇒ Rs. 10

Gain% = (Gain/C.P) × 100

⇒ (Rs. 10/Rs. 80) × 100

⇒ 12.5%

Total Profit percent of Aditya is 12.5%

Numerical Ability Test - 3 - Question 4

In the given questions, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer
I. x2 = 361
II. y = √361

Detailed Solution for Numerical Ability Test - 3 - Question 4

Given:

I. x2 = 361

II. y = √361

Calculation:

I. x2 = 361

⇒ x = 19 or x = – 19

II. y = √361

⇒ y = 19

Comparison between x and y (via Tabulation):

∴ We can clearly see that x ≤ y
Confusion Points
Both cases are different :
⇒ x2 = 361 → x = 19 & -19 (two values)
⇒ ​ y = √361→ y = 19 (only one value)
Note: Value coming out of perfect square root is always positive.

Numerical Ability Test - 3 - Question 5

Find the value of x.
(√16)3 × (√81)5 ÷ (27)2 ÷ (8)2 = 3x

Detailed Solution for Numerical Ability Test - 3 - Question 5

⇒ 16(3/2) × (81)(5/2) ÷ 272 ÷ (8)2 = 3x

⇒ 2(4 × (3/2)) × 3(4 × (5/2)) ÷ 3(3 × 2) ÷ (2)(3 × 2) = 3x

⇒ 26 × 310 ÷ 36 ÷ 26 = 3x

⇒ 3(10 – 6) = 3x

⇒ 34 = 3x

∴ x = 4

Numerical Ability Test - 3 - Question 6

What will come in place of the question mark ‘?’ in the following question?

40% of 265 + 93 × 4 – 4848 ÷ 24 × 8 = ?

Detailed Solution for Numerical Ability Test - 3 - Question 6

Concept used:
BODMAS rule

Given expression is,
⇒ 40% of 265 + 93 × 4 – 4848 ÷ 24 × 8 = ?
⇒ 106 + 729 × 4 – 4848 ÷ 24 × 8 = ?
⇒ 106 + 729 × 4 – 202 × 8 = ?
⇒ 106 + 2916 – 1616 = ?
⇒ 3022 – 1616 = ?
∴ ? = 1406

Numerical Ability Test - 3 - Question 7

What will come in the place of the question mark ‘?’ in the following question?

410 × ? × 26 = 67250 + 50010
Detailed Solution for Numerical Ability Test - 3 - Question 7

Follow BODMAS rule to solve this question, as per the order given below,

Step - 1 - Parts of an equation enclosed in the ‘BRACKETS’ must be solved first

Step - 2 - Any mathematical ‘OF’ or ‘EXPONENTS’ must be solved next

Step - 3 - Next the part of the equation that contains ‘DIVISION; and ‘MULTIPLICATION’ are calculated

Step - 4 - Last but not least, the parts of the equation that contains ‘ADDITION’ and ‘SUBTRACTION’ should be calculated

Now, the given expression:

⇒ 410 × ? × 26 = 67250 + 50010

⇒ 10660 × ? = 117260

Numerical Ability Test - 3 - Question 8

What will come in the place of the question mark ‘?’ in the following question?

Detailed Solution for Numerical Ability Test - 3 - Question 8

Follow BODMAS rule to solve this question, as per the order given below,

Step - 1 - Parts of an equation enclosed in the ‘BRACKETS’ must be solved first

Step - 2 - Any mathematical ‘OF’ or ‘EXPONEMTS’ must be solved next

Step - 3 - Next the part of the equation that contains ‘DIVISION; and ‘MULTIPLICATION’ are calculated

Step - 4 - Last but not least, the parts of the equation that contains ‘ADDITION’ and ‘SUBTRACTION’ should be calculated

Now, the given expression:

⇒ ? = 52

Hence, the required answer is 52.
Numerical Ability Test - 3 - Question 9

If the expenditure of Company B in 2008 was Rs. 220 lakh, what was its income in 2008?

Detailed Solution for Numerical Ability Test - 3 - Question 9

Profit% of Company B in 2008 = 35.
Let the income of Company B in 2008 be Rs. x lakh.
35 = [(x – 220)/220] × 100
∴ x = 297
Income of Company B in 2008 = Rs. 297 lakh.

Numerical Ability Test - 3 - Question 10

If the expenditures of Company A and B in 2006 were equal and the total income of the two Companies in 2006 was Rs. 342 lakh, what was the total profit of the two Companies together in 2006?

Detailed Solution for Numerical Ability Test - 3 - Question 10

Let the expenditures of each companies A and B in 2006 be Rs. x lakh.

And let the income of Company A in 2006 be Rs. z lakh.

So that the income of Company B in 2006 = Rs. (342 - z) lakh.

Then, for Company A we have:

45 = [(z - x)/x] × 100

45 = z/x – 1

x = z/1.45 ----(1)

Also, for Company B we have :

60 = [(342 – z – x))/x] × 100

1.6x = 342 – z

x = (342 – z)/1.6 ----(2)

Equating (1) and (2), we get :

1.6z = 1.45(342 – z)

1.6z = 495.9 – 1.45z

3.05z = 495.9

z = 162.59

Substitute the value of z in equation (1)

x = 112.13

Total expenditure of Companies A and B in 2006 = 2x = Rs. 224.26 lakh.

Total income of Companies A and B in 2006 = Rs. 342 lakh.

Total profit = Rs. (342 – 224.26) lakh = Rs. 117.74 lakh.

Numerical Ability Test - 3 - Question 11

If the incomes of Company A and B are equal in the year 2009, what is the ratio of the expenditure of A to that of B?

Detailed Solution for Numerical Ability Test - 3 - Question 11

%Profit = [(Income – Expenditure)/Expenditure] × 100

Let the incomes of each of the two Companies A and B in 2009 be Rs.x.

And let the expenditures of Companies A and B in 2009 be EA and EB respectively.

Then, for Company A we have :

25 = (x – EA)/EA × 100

EA = x/1.25

, for Company B we have :

30 = (x – EB)/EB × 100

EB = x/1.3

EA/EB = 26/25

Numerical Ability Test - 3 - Question 12

What is the percentage decrease in the percent profit of company A form the year 2007 to 2008?

Detailed Solution for Numerical Ability Test - 3 - Question 12

Percentage profit earned by A in year 2007 = 60%
Percentage profit earned by A in year 2008 = 40%
Decrease in percentage profit earned by A = (60 – 40)/60 = (100/3)%

Numerical Ability Test - 3 - Question 13
A train of length 450m travels with the speed of 50 m/s. The train crosses a man running in the same direction as the train in 10s. Find the speed of the man.
Detailed Solution for Numerical Ability Test - 3 - Question 13

Given:

Length of the train = 450m

Speed of the train = 50 m/s

Time taken to cross the man = 10 seconds

Formula used:

Speed(S) = Distance(D)/Time(T)

Calculation:

Let the speed of the man be x m/s

Speed at which train crossed the man = 450/10 m/s

⇒ 45m/s

In the same direction, the speed subtracts

⇒ (50 – x) m/s = 45 m/s

⇒ x = 5 m/s

The speed of the man is 5 m/s.

Numerical Ability Test - 3 - Question 14

What should come in place of question mark (?) in the following question?

(2222 ÷ 20) + (645 ÷ 25) + (3991 ÷ 26) = ?
Detailed Solution for Numerical Ability Test - 3 - Question 14

Follow BODMAS rule to solve this question as per order given below,

Step-1 - Parts of an equation enclosed in ‘Brackets’ must be solved first.

Step-2 - Any mathematical ‘Of’ or ‘Exponent’ must be solved next.

Step-3 - Next, the parts of the equation that contain ‘Division’ and ‘Multiplication’ are calculated.

Step-4 - Last but not least, the parts of the equation that contain ‘Addition’ and ‘Subtraction’ should be calculated.

Given expression:

(2222 ÷ 20) + (645 ÷ 25) + (3991 ÷ 26) = ?

⇒ 111.1 + 25.8 + 153.5

∴ ? = 290.4
Numerical Ability Test - 3 - Question 15

What will come in place of question mark (?) in the following equation?

182 - 2600 ÷ 52 = ? × 2

Detailed Solution for Numerical Ability Test - 3 - Question 15

Follow BODMAS rule to solve this question, as per the order given below,

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated

182 - 2600 ÷ 52 = ? × 2

⇒ 324 - 2600 ÷ 52 = ? × 2

⇒ 324 - 50 = ? × 2

⇒ 274 = ? × 2

∴ ? = 137

Numerical Ability Test - 3 - Question 16

What will come in place of question mark (?) in the following equation?
689 – [242 – 280 ÷ 5] = ?2

Detailed Solution for Numerical Ability Test - 3 - Question 16

Follow BODMAS rule to solve this question, as per the order given below,

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated

689 – [242 – 280 ÷ 5] = ?2

⇒ 689 – [242 – 56] = ?2

⇒ 689 – [576 – 56] = ?2

⇒ 689 – 520 = ?2

∴ 13 = ?

Numerical Ability Test - 3 - Question 17

If a sphere is inscribed in a cylinder, what would be the ratio between the volume of the cylinder and the sphere?

Detailed Solution for Numerical Ability Test - 3 - Question 17

Given:

A sphere is inscribed in a cylinder.

Concept:

When a sphere is inscribed in a cylinder.

Assumption:

Radius of the sphere = radius of the cylinder = r

Height of the cylinder = diameter of the sphere = 2r

Calculation:

∴ Volume of the cylinder

⇒ π × r2 × 2r

⇒ 2πr3

∴ Volume of the sphere,

⇒ 4πr3/3

∴ required ratio,

⇒ 2 ∶ (4/3)

⇒ 3 ∶ 2

Numerical Ability Test - 3 - Question 18

In the given question, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer

I) x2 = 784

II)

Detailed Solution for Numerical Ability Test - 3 - Question 18

As per the given data,

From I)

⇒ x2 = 784

⇒ x = ± 28

From II)

⇒ y = √(784) = 28

∴ From the above, we can say x ≤ y

Numerical Ability Test - 3 - Question 19

What will come in place of question mark (?) in the following question?

7 × 7 ÷ 7 + 7 ÷ 7 = ?
Detailed Solution for Numerical Ability Test - 3 - Question 19

⇒ ? = 7 + 1 = 8

∴ ? = 8
Numerical Ability Test - 3 - Question 20

What should come in place of the question mark ‘?’ in the following question?

69 + 123 × 2 - 4 = ? ÷ 3 + 11

Detailed Solution for Numerical Ability Test - 3 - Question 20

B- Bracket

O- Of

D- Division

M- Multiplication

A- Addition

S- Subtraction

According to this rule,

⇒ 69 + 123 × 2 - 4 = ? ÷ 3 + 11

⇒ 69 + 246 - 4 = ? ÷ 3 + 11

⇒ 315 - 4 = ? ÷ 3 + 11

⇒ 311 - 11 = ? ÷ 3

⇒ 300 × 3 = ?

∴ ? = 900

Numerical Ability Test - 3 - Question 21

Person X can complete a task in 12 days, and the ratio of the efficiency of X and Y is 3 : 2. X, Y, and Z working together can finish the task in 3 days. If they receive a total payment of Rs. 4800, what is the payment for Z.

Detailed Solution for Numerical Ability Test - 3 - Question 21

Calculation:

Let's denote the efficiency of X = 3a and the efficiency of Y = 2a

where 'a' = efficiencies.

Since, X complete the task = 12 days

Its total work = 12 × 3a = 36a.

X, Y, and Z together complete the task in 3 days

So their combined efficiency = 36a / 3 days = 12a.

Now, the ratio of the efficiency of X, Y and Z = 3 : 2 : Z

So, Z's efficiency = 12a - (3a + 2a) = 7a.

Now, the total payment is Rs. 4800

Z's share is proportional to his efficiency, which is 7a out of 12a.

Therefore, Z's payment is (7a / 12a) × Rs. 4800 = Rs. 2800.

Hence option number (C) is the right answer.

Numerical Ability Test - 3 - Question 22

Pramod bought 30 kg of rice at the rate of Rs. 8.50 per kg and 20 kg of rice at the rate of Rs. 9 per kg. He mixed the two. At what price (Approximate) per kg should he sell the mixture in order to get 20% loss?

Detailed Solution for Numerical Ability Test - 3 - Question 22

(9 - Price of mixture per kg)/(Price of mixture per kg - 8.50) = 30/20

⇒ 30x price of mixture per kg - 255 = 180 - 20x price of mix per kg.

⇒ Price of mixture per kg = (180 + 255)/50 = Rs. 8.70

S.P. of the mixture per kg = 8.70 × 80/100

⇒ Rs. 6.96 ≈ Rs. 7.00

∴ The selling price of the mixture is Rs. 7.00

Given

Pramod bought 30 kg rice price at = Rs. 8.50

Pramod bought 20 kg rice price at = Rs. 9.00

Mixture cost price

Cost price 30 kg = 30 × 8.50 = 255

Cost price 20 kg = 20 × 9.00 = 180

Cost price of the mixture = 255 + 180 = Rs. 435

Total bought rice is = 30 + 20 = 50 kg

Mixture per kg price is = 435/50

⇒ Rs. 8.70

S.P. of the mixture per kg = 8.70 × 80/100

⇒ Rs. 6.96 ≈ Rs. 7.00

∴ The selling price of the mixture is Rs. 7.00

Numerical Ability Test - 3 - Question 23
In a village election a candidate who got 25% of total votes polled was defeated by his rival by 350 votes. Assuming that there were only 2 candidates in the election, the total number of votes polled was?
Detailed Solution for Numerical Ability Test - 3 - Question 23

Given:

2 candidates in the election

A candidate who got 25% of total votes polled was defeated by 350 votes

Calculation:

Let the total number of votes polled = X

according to question,

75% of X - 25% of X = 350

⇒ 50% of X = 350

⇒ (50/100) × X = 350

⇒ X = 700

∴ Total number of votes polled 700.

Numerical Ability Test - 3 - Question 24

Find the ratio of the Number of Populations in the Year 2018 & 2020.

Detailed Solution for Numerical Ability Test - 3 - Question 24

From given line graph,

Total Population in 2019 = 120% of Total Population in 2018

= (120/100) *1,00,000 = 1,20,000

Total Population in 2020 = 140% of Total Population in 2019

= (140/100) *1,20,000 = 1,68,000

Total Population in 2021 = 160% of Total Population in 2020

= (160/100) *1,68,000 = 2,68,800

Total Population in 2022 = 180% of Total Population in 2021

= (180/100) *2,68,800 = 4,83,840

From above solution, Total Population in 2018 = 1,00,000

Total Population in 2020 = 1,68,000

ratio of Number of Population in Year 2018 & 2020 = 1,00,000 : 1,68,000 = 25 : 42

Numerical Ability Test - 3 - Question 25

Find the difference between the population in the Year 2021 & 2020.

Detailed Solution for Numerical Ability Test - 3 - Question 25

From given line graph,

Total Population in 2019 = 120% of Total Population in 2018

= (120/100) *1,00,000 = 1,20,000

Total Population in 2020 = 140% of Total Population in 2019

= (140/100) *1,20,000 = 1,68,000

Total Population in 2021 = 160% of Total Population in 2020

= (160/100) *1,68,000 = 2,68,800

Total Population in 2022 = 180% of Total Population in 2021

= (180/100) *2,68,800 = 4,83,840

From above solution, Total Population in 2021 = 2,68,800

Total Population in 2020 = 1,68,000

difference between population in Year 2021 & 2020 = 2,68,800 - 1,68,000 = 100800

Numerical Ability Test - 3 - Question 26

Find the average of the number of Populations in the year 2018 & 2019 together.

Detailed Solution for Numerical Ability Test - 3 - Question 26

From given line graph,

Total Population in 2019 = 120% of Total Population in 2018

= (120/100) *1,00,000 = 1,20,000

Total Population in 2020 = 140% of Total Population in 2019

= (140/100) *1,20,000 = 1,68,000

Total Population in 2021 = 160% of Total Population in 2020

= (160/100) *1,68,000 = 2,68,800

Total Population in 2022 = 180% of Total Population in 2021

= (180/100) *2,68,800 = 4,83,840

From above solution, Total Population in 2018 = 1,00,000

Total Population in 2019 = 1,20,000

average number of Population in year 2018 & 2019 together = (1,00,000 + 1,20,000)/ 2 = 1,10,000

Numerical Ability Test - 3 - Question 27

Find the percentage increase in the number of Populations of year 2020 with respect to the total population in 2018.

Detailed Solution for Numerical Ability Test - 3 - Question 27

From given line graph,

Total Population in 2019 = 120% of Total Population in 2018

= (120/100) *1,00,000 = 1,20,000

Total Population in 2020 = 140% of Total Population in 2019

= (140/100) *1,20,000 = 1,68,000

Total Population in 2021 = 160% of Total Population in 2020

= (160/100) *1,68,000 = 2,68,800

Total Population in 2022 = 180% of Total Population in 2021

= (180/100) *2,68,800 = 4,83,840

From above solution, Total Population in 2018 = 1,00,000

Total Population in 2020 = 1,68,000

Percentage increase in Population of year 2020 with respect to total population in 2018 = (1,68,000 - 1,00,000)/1,00,000 = 68%

Numerical Ability Test - 3 - Question 28

Find the Total Population in the year 2021 & 2022 together.

Detailed Solution for Numerical Ability Test - 3 - Question 28

From given line graph,

Total Population in 2019 = 120% of Total Population in 2018

= (120/100) *1,00,000 = 1,20,000

Total Population in 2020 = 140% of Total Population in 2019

= (140/100) *1,20,000 = 1,68,000

Total Population in 2021 = 160% of Total Population in 2020

= (160/100) *1,68,000 = 2,68,800

Total Population in 2022 = 180% of Total Population in 2021

= (180/100) *2,68,800 = 4,83,840

From above solution, Total Population in 2021 = 2,68,800

Total Population in 2022 = 4,83,840

Total Population in 2021 & 2022 together = 2,68,800 + 4,83,840 = 7,52,640

Numerical Ability Test - 3 - Question 29

What approximate value will come in the place of the question mark ‘?’ in the following question?
37.5% of 120.08 + 56 ÷ 7.99 = ?

Detailed Solution for Numerical Ability Test - 3 - Question 29

Given:
37.5% of 120.08 + 56 ÷ 7.99 = ?
Rule:

Calculation:

⇒ 37.5% of 120.08 + 56 ÷ 7.99 = ?

⇒ 45 + 56 ÷ 8 = ?

⇒ 45 + 7 =?

⇒ ? = 52

∴ The required answer is 52

Numerical Ability Test - 3 - Question 30

Three numbers are in the ratio 6 : 9 : 10. If the sum of three numbers is 100, then find the smallest number.

Detailed Solution for Numerical Ability Test - 3 - Question 30

Given:

The sum of the numbers = 100

Calculation:

Let the numbers be 6x, 9x, and 10x respectively

According to the question

⇒ 6x + 9x + 10x = 100

⇒ 25x = 100

⇒ x = 100/25 = 4

The smallest number = 6x = 6 × 4 = 24

∴ The required result will be 24.

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